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While I was trying to find out why the first stiefel whitney class of a vector bundle is zero iff its orientable, I figured out that $w_1(Bundle)=w_1(Top\,exterior\,power\,of\,bundle)$ which is zero if the vector bundle is orientable. I did this using the kunneth formula for the exterior power of a direct sum and the formula for the characteristic classes of a tensor product of line bundles.

Question: Given a real vector bundle $E \to B$, is $w_i(E)=w_1(\Lambda^i E)$?

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  • $\begingroup$ The equation you've written doesn't make sense, since $w^i\in H^i$ and $w_1\in H^1$. $\endgroup$ – Eric Wofsey Jun 23 '17 at 19:54
  • $\begingroup$ sorry :) can I delete this question? $\endgroup$ – user062295 Jun 23 '17 at 19:55
  • $\begingroup$ Given that $w_1(E) = w_1(\bigwedge^{\operatorname{rank}E}E)$, it seems that the natural question to ask is whether $w_1(E) = w_1(\bigwedge^i E)$. Is that what you meant to ask? $\endgroup$ – Michael Albanese Jun 24 '17 at 5:02

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